MA441: Algebraic Structures I

Lecture 23

1 December 2003

1Review from Lecture 22:

Deﬁnition:

A subgroup H of a group G is called a normal if aH =Ha for all a∈G.

We denote this by HCG.

Theorem 9.1: Normal Subgroup Test

A subgroup H of G is normal in G iﬀ

1xHx H for all x∈G.

2Deﬁnition:

For HCG, the set of left (or right) cosets of

H in G is a group called the quotient group

(or factor group) of G by H.

We denote this group by G/H.

Theorem 9.2:

Let HCG. The set G/H is a group under the

operation (aH)(bH)=abH.

3Example 11:

Let G=U(32)={1,3,5,7,...,27,29,31}.

LetH =U (32)={1,17}. (171 (mod 16))16

G/H is abelian of order 16/2=8.

In Chapter 11, we’ll learn about the Funda-

mental Theorem of Abelian Groups. This the-

orem tells us that an abelian group of order 8

must be isomorphic to Z , Z Z , or8 4 2

Z Z Z .2 2 2

Which of these three direct products is isomor-

phic to G/H?

4We will determine the elements of G/H and

their orders.

1H ={1,17}, 3H ={3,19}, 5H ={5,21}

7H ={7,23}, 9H ={9,25}, 11H ={11,27}

13H ={13,29}, 15H ={15,31}.

2We can rule out Z Z Z , because (3H) =2 2 2

9H, so |3H|4.

2 2Now (7H) = (9H) = H, so these are two

distinct elements with order 2. This rules out

Z , which has only one element of order 2.8

Therefore, U(32)/U (32) Z Z , which is16 4 2

isomorphic to U(16).

5Theorem 9.3:

Let G be a group with center Z(G). If G/Z(G)

is cyclic, then G is Abelian.

Proof:

Letg ∈GbesuchthatgZ(G)generatesG/Z(G).

For any a,b∈G, let

i iaZ(G)=(gZ(G)) =g Z(G),

and let

j jbZ(G)=(gZ(G)) =g Z(G).

6i jThen a=g x and b=g y, ∃x,y ∈Z(G).

iSince x,y commute with everything, and g

jcommutes with g , we see that

i j i+j j iab=(g x)(g y)=g xy =(g y)(g x)=ba.

Since the a,b were arbitrary, we have that G is

abelian.

7Theorem 9.4:

For any group G, G/Z(G)Inn(G).

Proof:

Deﬁne the map T :G/Z(G)→Inn(G) via

gZ(G)7→ .g

(We’ll use Gallian’s deﬁnition of inner auto-

1morphism: (x)=gxg .)g

We need to show that T is a well-deﬁned func-

tion, that is one-to-one, onto, and preserves

the group operation.

8ToshowthatT iswell-deﬁned,supposegZ(G)=

hZ(G). We’ll show that both cosets map to

the same inner automorphism.

1Now gZ(G) = hZ(G) implies h g ∈ Z(G).

1 1Then for all x∈G, h gx=xh g.

Bymultiplyingontheleftby handontheright

1 1 1by g , we have gxg =hxh .

Therefore = .g h

9One-to-one: reverse the argument. = g h

implies gZ(G)=hZ(G).

Onto: by deﬁnition of T. For any , the cosetg

gZ(G) is a preimage under T.

Group operation:

T(gZ(G)hZ(G))=T(ghZ(G))= .gh

T(gZ(G))T(hZ(G))= = .g h gh

1 1(Recall: (x)= (hxh )=(gh)x(gh) .)g gh

10